Question 8 of 8: Stepped Shaft in Torsion

Question 8: Stepped Shaft in Torsion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

G = 25 GPa; T_A=25 kN·m (CW), T_B=30 kN·m (CCW), T_C=35 kN·m (CW); shaft fixed at E.

Find. (a) Maximum shear stress and its radial variation. (b) Angle of twist of A relative to the fixed end E. (c) The effect of doubling all three torques.

Approach. Take clockwise as positive; find the internal torque in each of the four segments by summing the applied torques to the free (A) side of each cut, then apply $\tau=Tr/J$ and $\phi=TL/(GJ)$ segment by segment, summing the angles algebraically.

1. Internal torque in each segment (CW+). $$T_{AB}=T_A=+25, \quad T_{BC}=T_A+T_B=25-30=-5, \quad T_{CD}=T_{DE}=T_A+T_B+T_C =25-30+35=+30\ [\text{kN}\cdot\text{m}]$$
2. Polar moments of inertia. $$J_{AB}=\frac{\pi(100)^4}{32}=9.817\times10^6\ \text{mm}^4, \quad J_{BC}=\frac{\pi(60)^4}{32}=1.272\times10^6\ \text{mm}^4$$ $$J_{CD}=\frac{\pi(120)^4}{32}=20.36\times10^6\ \text{mm}^4, \quad J_{DE}=\frac{\pi(120^4-60^4)}{32}=19.09\times10^6\ \text{mm}^4$$
3. Part (a): shear stress in each segment, at its outer radius. $$\tau_{AB}=\frac{T_{AB}r_{AB}}{J_{AB}}=\frac{25\times10^6(50)}{9.817\times10^6} =\boxed{127.3\ \text{MPa}}\ \text{(governs)}$$ $$\tau_{BC}=117.9\ \text{MPa}, \qquad \tau_{CD}=88.4\ \text{MPa}, \qquad \tau_{DE}=94.3\ \text{MPa}$$ Segment AB governs at $\tau_{max}=127.3$ MPa ($<200$ MPa yield, OK). In every segment τ varies LINEARLY from zero at the shaft's centre to this maximum at the outer radius (for the hollow segment DE, from a non-zero value at the 30 mm inner radius up to the maximum at the 60 mm outer radius).
4. Part (b): angle of twist, A relative to E. Sum $\phi_i=T_iL_i/(GJ_i)$ (signed, CW+) over all four segments (G = 25{,}000 MPa): $$\phi_{AB}=+2.918^\circ, \ \phi_{BC}=-2.702^\circ, \ \phi_{CD}=+1.351^\circ, \ \phi_{DE}=+2.882^\circ$$ $$\boxed{\phi_{A/E}=\Sigma\phi_i = +4.45^\circ}\ \text{(net clockwise, viewed from A)}$$
5. Part (c): doubling all three torques. The shaft remains linear-elastic as long as no segment yields, so both τ and φ scale directly with T: every stress and the total twist would exactly DOUBLE (new $\tau_{max}=254.6$ MPa, new $\phi=8.90^\circ$). However $254.6\ \text{MPa} > 200\ \text{MPa}$ (yield), so segment AB (and possibly BC) would actually begin to YIELD; the true response would become elastic-plastic and the simple doubling relationship would no longer hold once yielding starts.

Final Results.